find the exact solution, using common logarithms, and a two-decimal-place approximation of the solution 4^x-3(4^-x)=8

\[\Large 4^x-3(4^{-x})=8\]first subtract 8 on both sides \[\Large 4^x-3(4^{-x})-8=0\]I would multiply both sides by 4^x \[\Large 4^{2x}-3-8(4^x) = 0\](think about exponent laws and you'll see how to get it.
Now it's like a quadratic.